Download When 1°, 2°, aromatic amines or aryl amines . (Rand

When 1°, 2°, or 3° amino group is directly attached to benzene nucleu�, compunds are known as aromatic
amines or aryl amines
1:2
�
.
Aniline, 1°
N-Alkylaniline
(2°amine)
(Aminobenz.ene)
(Rand R' may be any alkyl or aryl (group)
N, N-Dialkylaniline
(3° amine)
One should note that when amino group is indirectly attached to benzene ring, the compound is called as aryl
substituted aliphatic amine. In such compounds, -NH2 group behaves as that of aliphatic amines.
.
o-Toluidine
(o-Methylaniline)
o-Phenylenediamine
Aromatic amines
.
Benzylamine (1°)
N-Methylbenzylamine (2°)
Aryl substituted aliphatic amine
Most of the methods used for preparing aromatic amines · are very
1.
similar to that used for aliphatic amines.
Reduction of nitro compounds. The most widely used method for preparing aromatic amines is the reduction of
the nitro group to the amino group. This reduction can be achieved by catalytic hydrogenation, or most frequently
with an acid and a metal (Fe, Zn, Sn) or a metal salt like Sn02•
or.
(i)Fe,HO,(ii)OH. Aniline
Nitrobenzene
+
�
Q
Fe,Hcl
Q
-
OH
�
CH3
CH3
CH3
p-Nitrotoluene
p-Toluidine
�
(i)Sn,HCI
(ii) OHoacetophenone
m -Aminoacetophenone
m -Nitr
Remember that LiAIH4 does not reduce PhN02 to PhNH2.
Since nitro compounds are readily prepared by direct nitration, the primary aromatic amines obtained by the
reduction of these nitro compounds are readily converted into diazonium salts ; the diazonium group, in tum, can be
replaced by a large number of other groups. This route gives a common and most useful method for introducing different
groups in benzene nucleus.
Ar-H � Ar-N02 � Ar-NH2 � Ar-N2+ � Other compounds.
1.
Give steps involved in the following conversions.
Toluene to p-toluidine
(b)
o-Isop
(c)
m-Aminoacetophenone from benzene
(d)
4-Isopropyl-l, 3-benzenediamine from benzene.
(e)
Toluene top-aminobenzoic acid.
(a)
11murr· m
1
Pf
·
ropylaniline from benzene
I
-
Selective reduction of one nitro group of a dinitro compound can often be achieved by NH4HS or through the use
of carefully measured amount of hydrogen sulphide in aqueous (or alcoholic) ammonia
&
N02
I
I
t.
m
-Dinitrobenzene
NHi
NH4HSor
.61_
.
m
N02
-Nitroaniline
Hofmann degradation.
2.
C6H5CONH2
Br2/NaOH
C6H5NH2
Aniline
Benzamide
3.
Curtius reaction
4.
By acidic hydrolysis of isocyanides. C6H5NC
5.
By reduction of azob enzene .
I-12,Ni
imder
pressure
o-
6.
=
NAr
+
HCOOH
-0
Sn,NaOH
-----­
(mild reduction)
ArNH-NHAr
hydrochloride.
N(CH3)iHO
6
H2,Ni
under
pressure
Aniline
Hydrazobenzene
+
(i) SnCl2, H30 , (ii) OH­
(vigorous reduction)
Hofmann Martius rearrangement. When a. 3° or 2° aromatic amine
primary amine
(CH3)H
heat
�
N, N-Dimethylaniline
hydrochloride
7.
·
NH-NH
Az oben zen e
ArN
+
Ho
� C6H5NH2
Q
CH3
�
hydrochloride is heated,
it is converted into
-
OH
�
�' �'
CH3
CH3
Commercially, aniline is prepared by passing ammonia over chlorobenzene in presence of Cu20 as catalyst and
under pressure and at 200°C.
Like aliphatic secondary amines, aromatic secondary amines can be prepared by reducing the corresponding
isocyanide. C6H5NC
�
C6H5NHCH3.
Aryl amines give most of properties of aliphatic amines in exactly similar fasi
h on, viz. solubility, basicity,
alkylation, acetylation, benzoylation, carbylamine reaction, condensation with a ldehydes to form Schiff base, oxidation
etc. Two properties deserve special attention, viz. reaction with nitrous acid and electrophilic substitution.
1.
with nitrous acid. As is aliphatic amines, each class of aromatic amine yields a different kind of
product when treated with nitrous acid, HONO. Primary aromatic amines react with nitrous acid to yield arene
diazonium salts.
Reaction
ArNlfi
+ NaN02 + 2HX
1°Aromaticamine
�\
.
Ar-N
=
N x- + NaX
+
+
�o
Arenediazonium salt
Although arenediazonium salts are far more stable than aliphatic diazonium salts, especially at low temperatures
(0--5°C), they slowly decompose even at low temperature, hence always used immediat�ly after preparation.
(a)
Why arenediazonium ions are more stable than alkyldiazonium ions ? This is due to two factors.
Electron-release from the ortho and para positions of the ring stabilizes the arenediazonium
+
ions.
NaN: ;
(b)
O='
+
..
_
The arene cation, Ar+, is very difficult to be formed as compared to R+ from
RN+=
N.
Secondary amines, both aliphatic and aromatic, react with nitrous to form N-nitrosamines which shown
Libermann's nitroso reaction
C6H5NHCH3 + NaN02 +HO �
N-Methylaniline
C6H5N(NO)CH3
N-Nitroso-N-methylaniline
Tertiary aromatic amines undergo ring substitution discussed in electrophilic substitution.
2.
Electrophilic Substitution. Arylamines contain two functional groups, the amino group and the aromatic ring.
The reactivity of the amino group is affected by its aryl substituent, and the reactivity of the ring is affected by its
amino group. The same electron delocalization that reduces the basicity and the nucleophilicity of an arylamine
nitrogen increases the electron density in the aromatic ring and thus makes arylamines extremely reactive
+
+
towards electrophilic aromatic substitution.
N
NH2
We know that the -NH2, -NHR, and -NR2 groups are ortho, para�
H
directing and exceedingly powerful activating groups
in
electrophilic aromatic substitution. These effects are due to the
formation of the especially stable intermediate carbocations I and II, in
which every atom (except hydrogen) has a complete octet of electrons.
&
I
,,.,:;
Thus the chief problem encountered in electrophilic substitution with aromatlf amines is that
are hightly reactive. For example,
(a)
Q
I
y
I
H
they
y
II
In halogenation substitution tends to occur at every available ortho and para position and thus monohalogenated
product can't be prepared
·
+ Br2
(aq .)
-------+
Aniline
Br
*
9
Ois
Br
·
"
Nitric acid not only nitrates, but also oxidizes the highly reactive ring as well, with loss of much material as
dark-coloured tar. Furthermore, in the strongly acidic nitration medium, the amine is converted into anilinium
ion (-NH3+); substitution is thus controlled not by the -NH2 group but by the -NH3+ group which, because
+N�
6
of its positive charge, directs the entering group to the meta- position instead of ortho, and para-.
Cone. HN03'
+
-NHi gp. o, p-directoP.
-NH3 gp.
m
-director
However, all these difficulties are overcome by protecting the amino group by acetylation, with either acetyl
chloride or acetic anhydride. Acetylation (-NH2----+ NHCOCH3) converts -NH2 group to acetamido (-NHCOCH3)
group which is o, p-directing but lesser activating toward electrophilic aromatic substitution than the parent -NH2 group.
: N........_C�
O
/H
0
II
�ca
II
0
Aniline
(electron pair can delocaliz.e
only to benzene !ing making
o,
Acetanilide Resonance in acetanilid due to amide group
(note that electron pair on N can also delocaliz.e to
amide group, hence -NHCOCH3 gp. becomes
weak activator than the-N� group)
p-positions highly reactive)
Thus protection of the amino group of an arylamine moderates the reactivity of the -NH2 group and permits
nitration (or halogenation) of the ring to be achieved in the required o, p-positions. Another important feature of the N­
acetyl protecting group is that after it has served its purpose, it may be removed by hydrolysis (acidic or basic) liberating
the parent amino group (deprotection).
Bromination
NHCOCH3
�
�
Br2 I ClfsCOOH
(bromination step)
Aniline
-
+
(i) �O, H ; (ii) OH
+
(deprotection step)
�
Br
o-Bromoaniline
p-Bromoaniline
Nitration
'
+
(i) H20, H , heat
+
-
(ii) OH
N02
o-Nitroaniline
p-Nitroaniline
r
.
'
,
Sulphonation
,
, , ', �Aniline is usually sulphonated by "baking" the salt, anilinium hydrogen sulphate at 180-200°C ; the main
product is the para isomer,
·
•
·
-
+
N� HS04
6
Aniline
6_1_ao-_20_0 oe
180-200"C �
+
�
3hours
S03Sulphanilic
Sulphamic
Anilinium
hydrogen sulphate
Q
acid
acid
It is believed that sulphonation of amines proceeds by a mechanism that is entirely different from ordinary
aromatic substitution.
Sulphanilic acid (p-Aminobenzenesulphonic acid) is a salt of special type, called a dipolar ion* (sometimes called
a Zwitterion). It is formed by the reaction of on acidic group (-S020H) and a basic group (-NH2) that are part of the
same molecule, hence it is also called inner salt. It is interesting to note that its properties are different from that of a typical
amine and a sulphonic acid. Such properties are (a) high melting point, (b) insolubility in water and organic solvents, (c)
solubility in aqueous NaOH, and (d) insolubility in aqueous HO. These properties cari. be explained on the basis of its
structure.
(a)
(b)
(c)
High melting point is due to its ionic nature.
·
Insolubility in organic solvents is also due to its ionic nature. Its insolubility in water is typical of dipolar salts.
Not all salts dissolve in water.
Its solubility in aq. NaOH is because of transference of H+ from the weakly acidic -NH3 +group to strongly
basic OH- ion to form p-aminobe�ene sulphonate ion (Il), which, like most sodium salts, is soluble in water.
+
No reaction
HO
+---
Q
S03-
I (Insoluble in water)
II
�
S03(Soluble in water)
(d)
Its insolubility in aq. HO is due to the fact that the -503- ion is too weak base to accept H+ from strong acids
(recall that sulphonic acids are strong acids, hence their anions are very weak bases).
1.
Account for the fact that p-aminobenzoic acid, p-NH2C6H4COOH, does not exist as a dipolar ion, although 2-aminoethanoic acid
.
(glycine) exists as a dipolar ion..
Friedel-Crafts reactions are normally not successful to unprotected anilines.
���
COCH3
2-Ethylacetanilide
..
4-Acetamido-3-ethylacetophenone
A dipolar ion is formed only when a molecule contains both an amino as well as an acidic group, and the amine is more basic than the
anion of the acid.
·
·
Aniline having a strongly activating
salt (a weak electrophile).
Coupling reaction.
• benzenediazonium
ArN2
+
+
+
H
�
(-NH2)
Ar-N N --0-Ntti
=
p-Aminoazobenzene
A strongly
activated ring
A weak electrophile
group undergoes coupling reaction with
Nitrosation. In an electrophilic aromatic nitrosation, the attacking reagent is nitrosonium ion. The nitrosonium
ion is a very weak electrophile and hence nitrosation can take place only in rings bearing powerfully activating
dialkylamino
or hydroxy (-OH) group, i.e., in tert-amines and phenols.
(-NR2)
NO
p-Nitroso-N, N- dimethylaniline
N, N-Dimethylaniline
1.
Complete the following reaction by supplying structure to the bracketed compounds.
(a)
o:
NH2
+
Glyoxal
----+
(b)
[E]
NH2
NH2
(c)
2.
�H5 CHO
.
Aniline
[G)
o:�
lmolof
HON0,5°C
[F]
H o+
HNO
--.2+ [HJ+ [I] � [J] + [K] + [L]
Arrange the following amines in their decreasing basic character.
(a)
6 cS
�NH2
and
6
(b)
6
[]
N
and
0
H
When a primary aromatic amine dissolved or suspended in cold aqeuous mineral acid, is treated with sodium
nitrite, arenediazonium salts are formed.
+ NaN02+2HX
1° Aromatic amine
o-s•c
�
A diazonium salt
As described earlier, aryl diazonium ions are substantially more stable than alkyl diazonium ions, and are of
enormous synthetic value. The important synthetic reactions of diazonium salts may be divided into two classes. (a)
replacement reactions, in which nitrogen is lost as N2 and its place is taken by some other atom or group, and (b) coupling
reactions, iri which the nitrogen is retained in the product.
1.
-N02•
Reactions involving replacement of -N+ = N. Diazonium group can be replaced easily be any one of a number
of groups or atoms like -F, -Cl, -Br, -I, -CN, -OH, -H, and
Most of these replacement reactions do
not require any special and drastic condition ; simply the requisite reagent is added to the solution of diazonium
salt which is then gently warmed. The required substituted compound is formed along with the evolution of
nitrogen..
Sandmeyer reaction). Arenediazonium salts react with I
Replacement of the N2+group .by -Cl, -Br, or -CN
cuprous chloride, cuprous bromide, and cuprous cyanide to give products in which diazonium poup has been
iihe
-
JEP:laarl by- Cl,-�. and - CN �. � Jl'ft t i os me known as Sandmeyer reactions.
Ar-N2+x- � Ar-X
+
N2 (X
=
-Cl, ...,-Br, -or-CN)
Sometimes the synthesis is carried out by a modification known as the Gattermann reaction, in which copper
powder and hydrogen halide (HCl or HBr) are used in place of cuprous halide.
·
Replacement by
(b)
·
N0 2 group
-
HBF4
Ar-Ntx-
Ar-N2+BF4"'.'J..
� Ar-F + BF3 + N2
Replacement by -OH. Diazoniuin salts react with water, of course slowly at low temperature, even ice cold, and
(e)
vigorously at room or high temperature to yield phenols and this is the reason why these salts are used immediately
after preparation.
Ar-N2+x-
·
+
H20
�
Ar-OH+ N 2
+
H+
This is the most general method for the preparation of phenols. Sulphuric acid is normally used instead of
hydrochloric acid in the diazotization steps so as to
cationic intermediate; HS04- is less nucleophilic
minimise
than c1-.
the competition with water for capture of the
As we shall see later, the phenol so formed may undergo coupling reaction with the unreacted diazonium salt.
Since this coupling takes place easily in alkaline medium, we can minimize this further by making the solution
more acidic; for which diazonium solution is added slo wly to a large volume of boiling dil. H2S04•
Diazonium group can be replaced by -OH group also by adding cuprous oxide to a dilute solution of the
diazonium salt containing a large excess of cupric nitrate.
Ar-OH
This
variation of Sandmeyer reaction is a much simpler and safer procedure than the above older method for
preparing phenol.
(fJ
Replacement by hydrogen (deamination via diazotization). This can be brought about by a number of reducing
agents, e.g., H3P02, C2H50H, NaBH4' or Na2S03i m ost useful of these is hypophosphorous acid, H3P02.
Ar-N2+x-
+
H3P02
+
H20 ----+ Ar-H
+
H3P03
+
HX
Aniline can directly be converted to benzene by carrying out its diazotization in ptesence of hypophosphorous
acid for which amine is dissolved in hypophosphorous acid and sodium nitrite is added (diazonium salt is
reduced as fast as it is formed).
H3POi, NaN02
Ar-H + N2 + H20
.
Replacement of -NH2 group via the diazonium group by hydrogen is very
reaction because introduction
of -NH2 group (precursor of -N2X) can be used to introduce the new group in a p-po$i tion which is sometimes
not possible by �rect reaction as illustrated in the following preparation.
Ar -NH2
�ful
(g)
Replacement by phenyl group (Gattermann rtaction) :
H3C
-Q--
N cc
;
�::;:at,
H3C-O-O- cH3
hence
Preparation ohn-bromotoluene : Here the two o, p-directing groups are situated meta for each other,
neither bromiriation of toluene nor methylation of bromobenzne would yield the required m-bromotoluene.
QJBr
methylation
bromination
(Not formed)
However, this can be prepared from toluene in the following way.
�
6
Q
HN03
H:zS04
Q
+
Fe, H
Br2
�
�&
NHCOCH3
(-NHCOCH3 is mum stronger
o, p-director than-Oi3)
p-Toluidine
p-Nitrotoluene
(separated from o-isomer)
Toluene
(to reduce the activating
effect of -NH2)
NHi
N02
Q
(CH3COhO
�Br �&
c�
�
OH-, H:zO
NHC�
NaN02
HQ
+ N2 O
�
.
6-Br
H�2
�.·
.'
m
.
1.
In diazotisation of arylamines, excess of mineral add is used. Explain.
2.
Complete the following by supplying structures to the bracketed compounds.
-Bromotoluene
.- -
-
"
.«
.
'-.
N�
(a)
¢
NaNO"HBF4
s·c
3.
4.
c6
H2S04
heat
NaNOz
a;;Q
[BJ
[AJ
NaN02, HQ
s•c
NJ4HS
--+
[q
heat
�
heat
I
�
(b)
[A]
[BJ
N, N-Dimethylaniline
[DJ
t
[q
Give intermediate steps to carry out the following transformations.
(a)
Toluene to p-toluic add
(b)
Nitrobeltzene to m-bromophenol
(c)
p-Nitroaniline to 1, 2, 3-tribromobenzene
(d)
Benzene to 1, 3, 5-tribromobenzene
(e)
Benzenetom-bromochlorobenzene
(/)
Anilinetop-dinitrobenzene
(g)
C6H5NH2 to C6H5D.
Direct bromination of toluene gives a mixture of o- and p-bromotoluenes which are very difficult to separate, so devise
that
can
be used for converting toluene to o- and p-bromotoluenes.
a
scheme
Coupling reactions of diazonium salts. Like rutrosonium ion (+NO), arenediazonium ions (ArNt) are weak
electrophiles, and are thus ca pab le of attacking only very reactive rings. Hence the aromatic ring undergoing
attack by the diazonium ion must contain a powerfully electron-releasing group, generally -OH, -NR21 NHR, -or NH2• Substitution mainly ocCU.rs para to the activating group.
An azo rompound
(G-NRz,
-N!ii, -NHR,\
=
or-OH
(a)
J
Couplings between arenediazonium ca tion, and phenols take place most rapidly in
slighlty
alkaline solution.
In alkaline medium, an appreciable· amount of phenol is present as a phenoxide. ion which is more reactive
toward electrophilic substitution than the parent phenol.
0--N=NCC+ 0--
oH
p-Hydroxyazobenzene
However, the solution should not be too alkaline because in strongly alkaline medium the arenediazonium ion
exists in equilibrium with an un-ionized diazohydroxide and diazotate ; neither of these couple.
NaOH
NaOH
(b)
Ar-N
Ar-N=N-OH
=
N-O- Na+
Arenediazonium ion
Diazohydroxide
Diazotate ion
(couples)
(doesnotcouplef
(doesnot couple)
Couplings between arenediazonium cation and amines take place mostly in
0--N=N+CC O-035-0-N=N o-
slightly
acid1c solutions (pH
�7).
N(CTI3)i
+
+
N(�)2
___.,.
-0-
-03S
N=N
-0-
N(CTI3>i
Methyl orange
The higher acidity increases the amount of protonated amine which is unreactive toward electrophilic substitution
Amine
Aminium salt
(couples)
(does not couple)
Azo compounds are usually intensely coloured because the azo linkage
(-N
=
N-)
brings the two aromatic
rings into conjugation. This gives an extended system of delocalized 1t electrons and allows absorption of light
in the visible region. Because of their intense colours, many azo compounds are· extensively used as dyes. Azo
compounds undergo reduction in the following manner.
(i)
(ii)
A
Ar-N
-N
r
=
=
N-Ar'
N-,-Ar'
H
Sn,NaOH
H
Ar-N-N-Ar
A hydrazo compound
(mild reduction)
(i)SnCl2,H+ ;(ii)OH­
(vigorous reduction)
Azo compounds can exist in
cis and trans forms ; the former being .less stable due to steric strain.
Ar
Ar
"'N=/
Ar
""'
.
•.
cis-Azobenzene
.
N=N
°"'Ar
.tr1ns-Azobenzene
Reduction. Arenediazonium salts
with sodium
sulphite
are reduced to phenylhydrazine hydrochloride by means of Sn02 and HCl
or
.
Sna2,HCI
or
Na2S03
Phenylhydrazine hydrochloride
Benzenediazonium chloride
(a)
[ 1
D
5 °C
[AJ
C2H5Br
(b)
C6H6
(c)
(CH3 C0)20
C6H5NH2
AICl3
HN03
His04
[AJ
[ BJ
NBS
[BJ
CIS020H
[CJ
CuCI
�
[DJ
excess
NH3
NH3
.
p-N1trochlorobenzene
(i)H3 +o
[CJ
(ii) OH-
[DJ
(e)
�a
Solution:
(a)
Proceed backward with the structure of the knwon compound, p-nitrochlorobenzene and the knwon reagent.
+
-
cua
NaN02
Sandmeyer
HQ,5°
reaction
N02
[DJ
�CH3
(b)
6
[AJ
CJi2CH3
N02
[BJ
(Curtius reaction)
a
[C]
'¢ �·�:
.
Q
¢�m
[BJ isN3H I cone. H2S04
N�
rq
N�
[DJ
COOH
¢
a
[A]
(c)
H:Q-�
(d)
N=NC6lf5
[A]
C2fis
(e)
ff�
<JiN
[A]
Solution:
(a)
NaN02
HO,s0c
(i) HNO:i- �504
(b)
racemate
(ii) Sn, HO
(iii) Off
N�
'
resolve with
(+}-or(-}­
tartaric acid
�
h�c,H,
·v
NH2
enantiomers
separated
reacemate
(i) HONO, 5°C
(ii)H3P02
separated
sec-Butylbenzenes
(c)
6
6
HONO
5"C
Aniline
6
CuCN
©
HNO�
�4
6
�
Benzene
·•
0 ¢NC>i
(e)
(j)
©
6
Q
(ii) OH-
;6Co·<;::,EJ,
(QHONJ,S"C
(ii) CuCI
a
�
�4
(i)Sn,HO
N02
<;::,lJ, NH,
Al03
Benttne
�
CN
¢
(i) HONO, 5°C
(ii)CuCN
Q
a
�
;6Co<&fs
)S.,Ha;Olf
Q
(ii) HN02' 5"C
0
a
-
�
6:
�
Q) HMO,. a,so,
(ii)Fe,HO;OH
NHi
C�CH3
C�CH3
HON0,5"C
©l_
(i)HBF4
N2 0
(g)
©
(i) HN031 H.iS04
(ii) Fe, HO; OH-
Isopropylbenzene
Q
�N�
CiisCCXl
(l) HNO:y li.iSO\
(ii) OH
NHi
+
-
(ii) heat
Ethyl
©l_F
m -flu orophenyl ketone
Q
NHi
(Cumene)-
f
C:E·�NH2
CN
NHi
·
�
c��
Clis�COO
Ben7.0ic acid
HNO�
(i)Sn,HO
(ii) (ClisCO)zO .
Q
OH-
�
6
hydro�
NHC�
N�
(d)
COOH
CN
+
NHz
NH�
�2
·HONO,s0c
�N�
·.
N2
� N�
m-Isopropylniuobenzene
Solution:
Solubility of the compound Xin water, and its reaction with NaOH to form Y with the loss of chlorine indicates that
Xis the hydrochloric acid salt. Further compound Y (not having chlorine) seems to be 1° primary aromatic amine
due to its reaction with HCl (diazotisation) and �-naphthol (coupling) to form red precipitate. The 1° aromatic
amine hydrochloride should be C6H5NH3+a-, indicated by its N.E. 129.
C6H5NH3+a- +·H20 � C6HsNH2 + HzO + HCl
(X) NE
=
129
(Y)
C6H5NH3+a-+ NaOH � C6HsNH2 + HzO + NaCl
(X)
(Y)
Solution:
Since the compound (A}, C7H7N02 is insoluble in dil. ·acid and base, and its nitrogen is not removed on oxidation,
so it should contian -N02 group as one of the substitutents which is present as such in B. Hence the other
substituent in B (C7H5N04) should be -COOH as indicated by its solubility in aq. NaHC03. So, the compound
B,
C7H5N04 is C6H4(N02)COOH. The two substituents must be present in para to each other so as to explain the
formation of two isomeric monochloro substitution products. Thus (A) should be p-nitrotoluene
COOH
Two
possible
sites
: : '9
R
¢
vigorous
oxidation
N02
or
N02
(B)
¢
N02
(A)
Solution:
[A]
[B ]
*
(soluble in aq. NaHC03)
��®
One mononitro derivative
Students are advised to start the problem from any knwon reaction fact or strucutre given in the problem. For example,
in the present question it is given that compound [B] is soluble in aq. NaHC031 so it must contain -COOH group.
Further it is given that (B) forms only one mononitro derivative which indicates that it should contian two -COOH
groups and that too in para to each other.
COOH
COOH
�x
Substitution
¢
COOH
COOH
{B) All four positions
are
equivalent
Now since compound B is obtained by acidic -KMn04 oxidation of A, A should have two substituents and one
substituent must contain nitrogen in such form which is insoluble in acids and bases an.d is liable to be converted to
-COOH group by KMn04 + H2S04.
The nitrogen containing group which coincides with all the given* characteristics is -CONH2. Hence the other
substituent in A (C8�NO) should be -CH3 and that too in para position to explain the formation of B.
COOH
¢
COOH
{B)
Solution:
In such questions, first draw structures of both of amines
replace the two amino groups by azo linkage, -N
=
in such a way that their amino groups face each other. Now
N-.
(A)
(X)
The two possible pairs that can be used for preparing the azo compound X by coupling are
CH3
A
H�,'o+
Br
CH
Br
Q-rn, H0-0 ON,-0--CH,
or
+
(From amine A)
(From amine B)
However, the first combination is not feasible because the benzene ring does not bear reactive -OH or -NH2
group, i.e., it is not sufficiently reactive to be attacked
"
by
the electrophilic diazonium cation.
Other important N containing groups are -N02 (not removed by .KMn04 + Hi50�
ben7.ene ring).
,
NH2
-
,
-NHCOCH3'
etc. (liable to destroy the
Solution:
[A) C6H5N02.CJ{4
The above series of reactiodS lead to following points.
·
(i)
Compound (C) is a· diamine and is fonned by treating [B] with strong HCl, involving rearrangement, [C] must
have two
groups in para positions.
-NH2
Thus the compound [B] must be a substituted hydrazobenzene with undergoes rearrangement (benzidine
rearrangement) on treatment with strong mineral acid.
(ii)
Fonnation of hydrazobenzene by the reduction of compound A with Sn and OH- indicates that A must be
2-ethylnitrobenzene.
[ B]
[A]
Solution:
Due to the presence of Nin ring, pyridine behaves like a strongly deactivated benzene (e.g. nitrobenzene), thus positions
2 and 4 are deactivated towards electrophiles. This is evident by the fact that Friedel-Crafts reactions fail while other
electrophilic substitutions require unusually strong conditions. Thus formation of 3-substituted product (the relatively
less deactivated position) can be explained by higher stability of the corresponding intermediate.
Attack at 3-position :
�H
l � 'N02
>
(
)
__
N
+
Attack at 2-position :
� ·H
l!... �.J(NOi
,
.,., __
+� ·H
�N.J(NOi
(as well
4-) substituted product is less stable than the 3-substituted
.-c-->
�+-H
�N .J(N02
No octet on N (Unfavourable)
Thus the intermediate corresponding to
intermediate.
2-
as
o u
on:
This is an example of nucleophilic substitution, recall that pyridine is deactivated toward attack by electrophiles, it is
activated toward attack by nucleophiles. Thus if a good leaving group is present either at 2 or 4 position, a nucleophile
can attack and displace the leaving group. The intermediate due to attack at 2 or 4 position is stabilized by delocalization
of the negative charge, since this stabilization is not possible if attack occurs at the 3-position, 3-chloropyridine is not
converted to 3-methoxypyridine.
Nucleophilic attack at C-2
II
Nhasoctet,
III
charge on N
(especially stable)
-ve
I
Similar situation arises in case of 4-chloropyridine.
Nucleophilic attack at C-3 (not observed)
�Cl
IL �
N
None of the structure has -ve charge on N {hena! unfavouable situation)
base)I �
'4 N .J
:�HH
H2'Pt)I
CX):
H
\
I
Solution:
+
rl � H
� NA cOCH3
�E--,.)J
.C)<�
)
3
H
H
(
Every atom has a full octet
(especially stable)
The given reactions (solubility in aq. HCI, no gas with HONO and a precipitate with C6H5C02 Cl I NaOH)
indicate that the given compound is a secondary amine.
(ii)
Since the given compound eliminates nitrogen by two sequence of Hofmann degradation, N must be present in
the ring.
(iii)
FormatiQJl of two unbranched alkenes as the final products indicates that the alkyl group or groups consisting
of 3-C must be present on the a-carbons and it must be unbranched. Two possibilities arise : (a) one CH3 and one
Ci.Rs groups are present on the two a-carbons, (b) one n�ff,. group is present on one of the a-carbons. However,
YCH,CH,OI,
(1)2CH3 I
(ii) Ag20, heat
-
Coniine(X]
H
2-Ethyl-6-methyl
piperidine, isomer of (X)
(i)2CH3I
(ii)Ag20,�
N
.......
H�
2-n -Propylpiperidine,
-�
�
[Y]
(i) CH3f
(ii) Ag20, heat
p
1, 4-octadiene
CH3
f).,
/ '
H3C
CH3
�
(i)CH�
(ii) Ag20, heat
+
�
1, 5-octadiene
� +�
1, 6-0ctadiene
l, 5-0ctadiene
(MCQ - ONE option correct)
1.
Which of the following is not a property of sulphanilic acid ?
It is soluble in aq N a H
(a)
(b) It is soluble in aq. HO
It is insoluble in organic solvents
(c)
It does not melt but decomposes.
O
.
6.
Predict the product [BJ in the following series of reactions.
C
6H5NH2
(a)
(d)
CH03
�
[A]
(b)
C6H5CH2NH2
C-o-
H3
N
=
N
-o-
NMe2 can be produced
7.
by coupling reaction of which of the following pair?
(a)
�C
(b)
H3C
-O-
-O
NH2+
+H2N
0-
-o-
NMe2
(d)
C6H5NHCH3
(c)
How many diazo group will be introduced when resorcinol is
treated with excess of benzenediazonium chloride in alkaline
medium?
2
(b)
1
(a)
(c)
Both (a) and (b)
Neither of the two.
Arrange the following compounds in decreasing order of coupling
with benzenediazonium chloride.
(d)
+
6 6 6' 6'
(II)
(a)
(b)
IV
II
II
II
(c)
4.
>
>
>
I
II
>
I > IV > III
IV > I > III
IIl > IV > I.
Qr
Q Q
c�
(I)
c�
=\:)=
6
(a)
o
(c)
H�-o-CHO(d)
(b)
a�a
�
(a)
I
(c)
(IV)
10.
(a)
(c)
Main product
(b)
O
0
.
(d)
No reaction.
Benzylarnine
(b)
Aniline
Acetanilide
(d)
p-Nitroaniline
Which of the foll owing is stronger than aniline ?
p
NHCOCHs
�-0-�
-Q-
-------t
Which of the following has lowest pKb value?
(a)
H1N
�
NH 2 + 21Cl
N02
11.
-------t
.
-
What should be the final product in the following reaction ?
N02
NCJi
o
a mixture of (a) and (b)
m Nitroaniline
NO,
(a)
l > Il > III > IV
(b)
IV > III > Il > I
IV > IIl > l > II
(c)
(d) II > I > III > IV.
Predict the nature of P in the following reaction.
�
(c)
(d)
-
02
N 02
(III)
+
p-Nitroaniline
+
N02
(II)
o Nitroaniline
(a)
(b)
III
+
6
Nil.
X. Here Xis mainly
Cone. H2S04
----->
(IV)
(III)
>
9.
(d)
Arrange the following diazonium cation in decreasing order of
coupling with phenol.
+
5.
>
Conc.HN03
8.
·
(I)
(d)
3
C6H5NH2.
NMe2
(c)
3.
[B]
C6H5COOH
'
2.
tt3o+
----'-----+
cooH
6
r
�CH3
(c)
(b)
(d)
None.
The increasing order of pKb values for the three anilines (I, Il and
ID) is
16.
I>IV>ill>Il
I > II > IV > III
(d) IV > I> m>II.
Which of the following is most basic, and which on is least ?
(c)
1'lli2
6(
n
13.
I
(a)
(c)
<
I
<
n
Ill
<
III
II
<
III
II
n
<
<
I
<
<
I
Ill.
Arrange the following amines according to their decreasing Kb
values.
¢¢¢¢
14.
(a)
(b)
(c)
(d)
CH3
OCH3
I
II
18.
19.
I
> II > III > IV
IV > III > II > I
II > I > III > IV
II > I > IV > Ill.
20.
Of the four orders given below for the basic character of the four
compounds, which one is correct order ?
6
�
N
21.
N
m
II
15.
17.
(a)
(b)
(c)
(d)
IV
IV > III > II > I
IV > I > II > III
IV > II > I > III
I > II > IV > III.
The correct order for decreasing basic character ofthe four amines
I to IV is
22.
23.
n
mH
�-
m
Il and I respectively
0
IV
(a)
(b)
(c)
IV and I respectively.
I and IT respectively (d)
When a 2°aromatic amine hydrochloride is heated, it leads to the
formation of 1° amine hydrochloride, the reaction is known as
Hofmann degradation
Fries migration
Hofmann Martius rearrangement
None ofthe three.
Fluorobenzene can be prepared from aniline via diazotisation,
the reaction is known as
(a)
Sandmeyer reaction
(b) Gattermann reaction
(c)
Schiemann reaction
(d) ModifiedSandmeyerreaction
Benzenediazonium chloride on reaction with phenol in weakly
basic medium gives
(a) diphenyl ether
(b) p-hydroxyazobenzene
(c)
chlorobenzene
(d) benzene.
A positive carbylamine test is given by
(a) N, N-dimethylaniline
(b) 2,4-dimethylaniline
( c)
N-methyl-o-methylaniline
(d)
p-methylbenzyl amine
Which of the following statement is not correct regarding aniline?
(a)
It is less basic than ethylamine
(b) It can be steam distilled
(c)
It reacts with sodium to give hydrogen
(d) It is soluble in water
Benzenediazonium chloride on reaction with phenol in weakly
basic medium gives
(a)
diphenyl ether
(b) p-hydroxyazobenzene
(c)
dtlorobenz.ene
( d)
benzene
The correct stability order of the following resonance structures is
(a)
(b)
(c)
(d)
+
H2C-N=N
(II)
-
©()NH
I
N
-
n
IT and IV respectively
I
(b)
(d)
OC
OCNH OCNK
NO2
+
Hi� -N=N
H2C-N�N
(Ill)
IV
(a)
(c)
I>II > IV>ID
II>I > ID>IV
choice questions with
(c)
LlAIH4
C6H Nc�
5
(d)
r - H C NH
Bll/NaOH
-e> 5 O
2
(b)
(d)
(IV)
ID > II>IV
ID>I>IV>II
I>
-20.21 (MCQ 1 or >1 option correct, Passage based, Matching, AIR)
DIRECTIONS for
Q. 1 to Q. 11 : Multiple
one or more th� one correct option(s).
1.
Which ofthe following reaction can be used for preparing aniline?
(a)
2.
Examine the following two structures for the anilinium ion,
predict which of the followin� st�tement is FALSE regarding the
two canonical structures for f!nilini um ion ?
Which of the following statement is true regarding reaction of p­
aminophenol with arenediazonium chloride?
+
6�
Ho-Q-Nlii+ArN;a---+
5
(I)
(a)
(b)
(c)
(d)
(Il)
(a)
II is not an acceptable canonical structure because
(b)
II is not acceptable canonical structure because nitrogen
(c)
II is not acceptable canonical structure because it is non-
(d)
II is acceptable structure.
carbocations are less stable than ammonium ions.
3.
aromatic.
8.
Sandmeyer reaction (b)
NaHC03
AgN03
(d) Carbylamine test.
Oxidation of A gives p-benzoquinone. A can be :
p-Aminobenzenesup
l honic acid
p-Aminobenzoic acid
(a)
M�
Aminoacetic acid
OH
Alanine
Which of the following pairs show coupling reaction?
(a)
90fa
6+6
10.
(c)
+
N�
(b)
+
(c}
(d)
NOi
(d)
Diazotised sulphanilic acid+ Dimethylaniline
Which of the following statement is false regarding following
reaction?
Cl
:¢(� +NH,
heat
pressure
Cl
(a)
No reaction is possible because-Cl is present on benzene
ring.
(b)
(c)
(d)
A nucleophilic substitution will take place in which both
-Cl will be replaced by two -:NHi groups.
A nucleophilic substitution will take place in which only
-Cl attached on C1 will be replaced by-N�.
A nucleophilic substitution will take place in which only
-Cl attached. on C4 will be replaced by-NH2•
Libermann' s nitroso reaction is used for testing
(a)
(c)
1° amine
phenol
pounds, the correct ones are
the rate of nitration of benzene is almost the same as that of
hexadeutrobenzene.
(b) the rate of nitration of toluene is greater than that of benzene.
the rate of nitration of benzene is greater than that of
(c)
hexadeutrobenzene
(d)
nitration is an electrophilic substitution reaction.
When nitrobenzene is treated with Br2 in presence of FeBr3' the
major product formed is m-bromonitrobenzene. Statements which
are related to obtain the m-isomer are
(a)
The electron density on meta carbon is more than that on
ortho and para positions
(a)
-
�
Among the following statements on the nitration of aromatic
com
11.
.
(d)
(c)
(b)
6.
p-Otloranilineandanilinehydrodlloridecan'tbedistinguishedby
(a)
(c)
9.
N2+CI-
5.
·
Which of the following can exist as inner salt ?
(a)
(b)
(c)
(d)
4.
has 10 valence electrons.
6
Reaction takes place at position 2 in presence of HCI.
Reaction takes place at position 3 in presence of NaOH.
No reaction occurs.
Four azo groups can be introduced in the molecule.
(b)
(d)
2° amine
3° amine.
The intermediate carbonium ion formed after initial attack
of Br+ at the meta position is least destabilised
Loss of aromaticity when Br+ attacks at the ortho and para
positions and not at meta position
Easier loss of W to regain aromaticity from the meta
position than from ortho and para positions.
INSTRUCTION for Q. 12to19: Read the passages given below and
th� questions that follow.
answer
Which of the three resonating structures is especially stable?
(b) II
(a)
I
(c)
III
(d) none
13. The unstability of the corresponding intermediate from
3-chloropyridine is because of the fact that
the intermediate has lesser number of canonical stru�
(a)
none
of the canonical structure has negative charge on N
)
(b
both of the above reasons
(c)
(d) none of the two
14. What would happen if 2-chloropyridine is replaced by 4-_
chloropyridine?
(a)
2-Methylpyridinewill beformed
(b) 3-Methoxypyridine will beformed
4-Methoxypyridine will beformed
(c)
(d) No reaction
18.
HD.3S
(d)
All are feasible
(a)
(c)
(d)
Neitheriscorrect
+
Ar-N=N OH
NaOH
H
I
16.
C6H5NH2 � C6H5N2x-
H20
>
+
(a)
(b).
·
use
III
(d)
(a)
(b)
N/O
N2+0-
+
+
Fe02
Zn02
(b)
Fe[Fe(CN)6]
Kz�[Fe(CN)6h
(d)
J1'ill 1!§1 Piii I .
I
WIWL I
IR
A11-.-.1Jm 'i!Ullllll lil!ililll'll
pKa of ammonium ions
@-Ntti
(a)
0.4
(b)
4.63
(c)
5.25
(d)
9.26
(e)
11.27
(A) Aldehyde+ Zn(Hg)
(a)
Carbylamine
(b)
distinguish between
primary and
other amines
White precipitate
(c)
Hydrocarbon
(B)
(C)
of H:z504 is
n Iii!
Basic compound
(A)
24.
NOi
(b)
Column-II
(E)
oH
Nao
Na2S
Column-I
HS04 is less nucleophilic than o-
0- � 0o0-
III
(d)
(d)
.
_
II and III
(b)
UtH
23.
(D)
·
H
II
+
Ar-N-N-ONa
Instructions for Q. 23 to 25 : Following questions are Multiple
Matching type Questions :
sulphuric acid is a stronger acid
(c)
sulphuric acid absorbs water
_
,(d) None of the three
17. Coupling reaction is not in feasib le in
+
Which of the above can undergo coupling reaction?
(b) II
(a)
I
C6HsOH+N2 +H+
In the preparation of the above diazonium salt,
preferred to HO because
NaOH
Ar-N•N-OH
I
because
(a)
resonance stabilization of the corresponding cation
(b) the arene cation, Ar+ is very difficult to be formecd
(c) both the above factors
(d) none of the two
NMe2--+
.
Coupling between arenediazonium cation and ammes take
place in strongly acidic conditions
Coupling between arenediazonium cation and phenols takes
place in slightly alkaline medium
Both are correct
The compound Xis
(a) NaN03
(c) NazS04
The compound Y is
(a) MgOz
(c) Fe03
The compound Z is
(a) Mg2[Fe(CN)�
(c) Fe4[Fe(CN)�3
Arenediazonium ions are more stable than alkyldiazonium ions
O-
In strongly alkaline medium, the arenediazonium cation exists in
the following equilibrium
(c)
15.
N2+0-+
Which of the following statement is false?
(b)
19.
-o-
(c)
NH3
c�
CNH
©
N
Column-I
+cone.HO
(B)
(C)
(D)
reaction Schiff bases
2, 4, 6-tri bromoaniline
(d)
Column-II
Antioxidation
Iii{
Statement-1: Anilinium chloride is more acidic than ammonium
chloride.
(A)
�:HN, em.-\�
a
a
(B)
@-
NHz+C6ffsN2
@-
+
(b)
29.
Curtius reaction
/
OH+ C6H5N
(c) Sulphanilic acid
_
U. H
STATEMENT-2 (Reason). Each question has 5 choices (a),
31 .
ti
26.
(b),
Statement 2 : Direct nitration of aniline gives a substantial
(c),
32.
33.
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a
Statement 1: Diazonium salts of aromatic amines are more
Statement 2: Diazonium ion shows resonance.
Statement 1: Ethylamine is soluble in water, whereas aniline is
not.
correct explanation for Statement-!.
Statement 2: Ethylamine forms hydrogen bonds with water
Statement -1 is True, Statement-2 is False.
Statement -1 is False, Statement-2 is True.
34.
Statement -1 is False, Statement-2 is False.
-
amount of p-nitroaniline
stable than those of aliphatic amines.
explanation for Statement-1.
alilil rn�rn --
Statement 1 : Amino group is o- and p-directing in aromatic
electrophilic substitution reactions, aniline on nitration gives a
substantial amount of m-nitroaniline.
Statement-1 is True, Statement-2 is True;Statement-2 is acorrect
�'1'/fild_X"l;"i'.ll!ii/l{!�'JL-'W.
HCI at
Statement 2 : Higher pKb means less basic.
(d) pH 9- 10
Methyl orange
(d) and (e) out of which ONLY ONE is correct
(e)
I
OOC followed by coupling with �-naphthol gives a dark blue
Statement 1: pKb for aniline is more than that of methylamine.
Note: Each question contains STATEMENT-1 (Assertion) and
(d)
.
STATEMENT -1: Aniline on reaction with NaN02
with J>-naphthol is due to the extended conjugation.
Instructions for Q. 26 to 35: Following questions are Assertion and
(c)
Nitrobenzene easily undergoes electrophilic
:
reaction of aniline with NaN02 /HCI at 0°C followed by coupling
Reasoning Type Q uestions:
(b)
Statement-2
is prepared by Friedel
STATEMENT- 2: The colour of the compound formed in the
�lllfil.lil�l1M-4U-l!il!ili Am\!lliliLJUll�li-�llfllU&oiltmll lil Lilll!ISl!ll-. _
(a)
p-02N-C2H5COCH3
precipitate.
Azo dye.
(D)
Statement-1 :
Craft's acylation of nitrObenzene.
substitution reaction
Azo dye.
---
(C)
28.
pH5-7
(a)
Statement-2 : Anilinium ion is resonance-stabilized.
II II I
.
W/J/flt!f;flif/l/j
Statement-1 : Benzene diazonium chloride does not give tests
35.
for nitrogen.
molecules.
Statement 1: Aniline undergo Friedel-Crafts reaction.
Statement 2: Aniline forms salt with aluminium chloride (a
lewis acid) which is used as a catalyst.
.
Statement 1 : Gabriel phthalimide synthesis is preferred for
synthesising aliphatic primacy amines.
Statement-2: Loss of N2 gas takes place during heating.
Statement 2 : Aryl halides undergo nucleophilic substitution
with the anion formed by phthalimide.
1.
Write
do� structures of the follo�� �ompounds.
(a) p-Tolwdine
(c) N, N-Die ylaniline
(e) Benzylanune
�
2.
3.
(g) p, p'-Diaminodiphenyl
(b)
(d)
(j)
(h)
I�
Anilinium chloride
Ifl
Diphenylamine
N- n-Propylaniline.
Co�pare the ��aviour of aru
. "line, N-methylaailine and N,
-
Give structures and names of the principal organic products
expected from the action of sodium nitrite and hydrochloric
acid on each of the following compound.
(a)
(c)
4.
(e)
o-Toluidine
N, N-Dimethylaniline
Benzylamine
(b)
(d)
4, 4'-Diaminodiphenyl
if)
Sulphanilic acid.
N-Methylaniline
Write balanced equations for the following reactions.
(a)
(b)
Benzanilide + aq. _NaOH (boiling)
Methyl formate +aniline
(d)
(e)
�
..
2, 4-Dimethylaniline
N-dimethylaniline tow ard each of the following reagents.
.
(b) excess of CH3I
(a) Dil. HO·
(c) C6H5COO+pyridine (d) C6H5S02Cl+aq.KOH
(e) Bromine water.
(c)
1i/j·
_,
(j)
5.
(g)
m-Nitro-N-methylaniline + HONO
m-Toluidine + aq. Br2 (excess)
p Toluidine+aq. Br2 (excess)
-
p-CH3C6H4NHCOCH3+HN03+H2S04
Benzanilide +Br2 +Fe.
Complete the following.
�
[AJ NH3 [BJ
J'iOs
ii
'Jj
(a)
Ci;H5COOH
(b)
C6H5NH2+[EJ+ [FJ � C6H5NHCONHC6H5
I
(c)
2, 4-Dinitroaniline
(d)
Aniline
I
IJ
I
I
fl
IJ
II
IJ
B12
water
NaNOz,HCI
s•c
[CJ�
[G �
J
[HJ
I (i) NaN02, HCI, s•c
[ J
(ii)C2H50H
[Jl
[DJ
©
�
9l3
1.
(a)
Conc.HN03
Conc.HiSQ4
(%)iCHCI, Al03
or
¢
(i)Sn,HO
(ii) OH
NH2
NOi
Toluene
(b)©
9l3
�CH-� AlClg
p-Toluidlne
c;:H{CH3).z
CH(CHg).z
6
�
~
(i)Sn,HQ
(ii)NaOH
�
�
o-Isopropylaniline
COCH3
(i)Sn,HO
)
(ii)NaOH
(d)©
� AIClg
(ii) Cone. HN03/Conc. �4
1.
(i)Sn,HO
(ii)NaOH
>
COOH
J¥04)
HN03
� �
I<Mn04
Remember that side chain is oxidized
1.
f
0
NOi
Di
�
(<)6
•
·
NH2
•-Aminoacetopheno
Acetophenone
(i) CH3CH
A0
�
when a deactivating group (N02) rather than an activating group CNH:z) is attached to ring.
H+
In p-aminobenzoic acid, -COOH is too · weakly acidic to transfer
to the weakly basic -NH:z group attached tO .the electron­
withdrawing benzene ring (remember that-SOJI is sufficiently strong acid to do so). However, aliphatic amino group (as in glycine)
is sufficiently basic to accept on
from the ---COOH group.
H+
(a) ��CH
6< .
�
I
[EJ Quinoxaline
[G) Schiff base
d N02
Y-NRz
2.
(a)
+
�
O
NH2
[KJ
c5
N
CHiNtti
>
-
0
(b)
>
Electron pair on
N is not a part of sextet,
hence localiz.ed
-0
=NI-iz
6
Electron pair
is effected by
effect of C6H5
Electron pair
on localiz.ed
OHC
[LJ
m
>
+
6
Electron pair
delocalized
I
�
lQJ
>
Electron pair
delocalized
Electron pair is a part of sextet and
hence delocalized and less
available for protonation
Acid prevents the coupling reaction by converting ArNH2 to its salt, ArNH3+ x-
1.
Unreacted aniline
Azocompund
M� - � � � � � �
6
� '"���� - ��€H
+ N O
N02
2.
-0-
N=N
(b)
3H
2 BF4
[AJ
[BJ
[q
Methyl orange
[DJ
[AJ
[BJ
[q
CH3
3.
(i) HN03/H2S04
(<)
CuCN
(ii) Fe,H
Toluene
2
2 a
p-Toluic add
�
N�
&
&�
6
©:�&
+
(b)
�
Fe
Nitrobenzene
�
Br
Br
HO
)
Br
Br
m -Bromophenol
Br
(i)Sn+HO
Br�
Br
1, 2, 3- Tribromobenzene
NMe2
+-
(d)
&
&
�
�
6 � · � 1¢r �
�
©
(e)
(i)�/Pt
�
©
Br
1, 3, 5-Tribromobenzene
Br
Br
Benz.ene
(ii) NaNO;i. HO
Fe
Benzene
N2+0NaN
0;1
HO
m -Bromodtlorobenzene
NaN0
O
02
o-Bromotoluene
·�
2
NaHCO;i. CufJ
p-Dinitrobenzene
Difficult to sepante
p-Bromotoluene
(b.p. 185°C)
(b.p. 182°C)
Hence following scheme should be opted.
¢
©:=
The two isomers are separated, each isomer is treated separately.
CH
3
(i)Fe,HO
(ii) NaNOi, HBr
¢�+ -
¢
CuBr
---+
N2 Br
N02
p-Nitrotoluene
Br
p-Bromotoluene
-20.1)
1
6
11
16
21
2
7 -
12
17
22
23
3
8
13
18
4
9
14
19
s
10
15
20
1.
In the Zwitterion of sulphanilic acid, the -so3- is too weakly basic (its conjugate acid -S03H is a strong acid) to accept H+ from
2.
Arenediazonium cation is a weak electrophile, hence it can couple only with those compounds which bear powerful electron-releasing
strong acids, hence it does not dissolve in HO.
group like-OH,
cation.
NH:z,
-
NHR, or�NR2. Hence toluene, having weak electron-releasing-CH3 group can't couple with diazonium
-
reaction in which,
Coupling reaction between arenediazonium cation and an aromatic compound is an electrophilic substitution
c6H5N2+,a weak electrophile reacts with the benzene compound having highly activating group. More is the activating (electron­
releasing) character of the group,easier will be coupling. Hence
4.
-NH2 > -OH > -NH3+.
make arenediazonium cation more electron-deficient,i.e., strong electrophile,while electron­
(like-N02)
Electom-withdrawing groups
5.
C6H5N02 oxidizes CH3 of p-NH2C6H4CH3 to CHO and itself reduced to C�sNH2•
-a-
>
releasing group makes diazonium cation a weak electrophile.
6.
7.
Resorcinol has one para and two ortho positions free .
8.
Aniline in presence of acids undergoes protonation to form C6HsNH3+ in which -NH3+group ism-directing.
(Q)-Ntti ::�: (Q)-Nii3
)
9.
Iodine chloride is an electrophilic reagent and because of high electronegativity of 0 it gives rise to 1+ and a-. Hence
OzN
10.
o-�+
N�
�
--0--NH,
+
ICI
-
Lower the pKb value of an amine,more is its basic character.
.. �
NHC�
(a)
6
(c)
(d)
11.
12.
e
n
pair: is delocalized onl_Y.,!I>
ring. but -I effect of�
also decreases buidty
I
pair is delocalized
e
only to ring
e palr is delocalized to ring .
as well as
gp. ; -I eHe<.t
to-�
of-Naz further � basi.city
Since higher the Kb value of an amine, lower is pKb value; thus the pK,; values of the three anilines will be of the order : I
13.
�OCH3
>
II
� increases basic
character due to
+ M effed of the-OCH3
gp.
QCH3
I
-cH3 increases
basic character
due to + I effect
ofthe�gp.
>
QCF3
m
-CF3 dec:relllleS
basic character
due to-I effect
of the-c:F3 gp.
>
Q
N02
IV
-NC>i decreases basic
character due to
-M and -I effects
<if the-N02 gp.
<
II
<
ill.
<;:Ha..
6
:�
14.
6
>
N
-nrnt
N
e paironNof the
;
N of the ring
also e pairondelocalized to
pairof Nof the
m
is delocalized
to ring
is not involved in sextet
ring not involv ed in
+I effectof-Cffs
e
©
>
I
II
· g
not involv ed in
Q
>
e paironN
epaironN
sextet;
further increases e
density
:�
15.
>
NH
�
pair localized
16.
18.
1 9.
localized
6
�
¢
0- \
N Cf
+
�
�
Q-oH
OH-
N
g-N .. -0(a,c,d)
8
(a,b,d)
9
7
PASSAGE I
12
(a )
13
(b)
15
(c)
16
(b)
14
PASSAGE2
20
(d)
21
(c)
22
26
(a)
27
{b)
32
(a )
33
(a)
MATCH
THE
FOLLOWING
AIR
5.
3
2
(a,b,d)
(a,b,d)
OPTION
4.
1
23
24
25
17
0H
(a,c,d)
4
(b,c,d)
5
(a,b,d)
(a, b, c)
10
(a,b,d)
11
(a, d)
(c)
18
(a)
19
(d)
(b)
{b)
7.
8.
6
(b,c)
31
(c)
(A) - b ; lB) - d ; (C) - e ; cm - a; ffi) - c
(A)- (c); lB)- (a); ca- (d); ID) - lb)
(A) - lb), (B) - a), (C) - (d), (0) - cl
28
34
(d)
29
35
(d)
(d)
30
(c)
(a)
Isonitriles. (C6H5NC) on reduction give 2° amines (C61f5NHCH3). AU Other three methods give aniline.
Nitrogen (N) has nod orbital, hence it can't have more than 8 electorns in the outermost shell. All other three options are false.
In p-NH2C6H4COOH, -COOH group is very weak so it can't transfer tt+ to the weakly basic amino group. All other three form
zwitterions.
In C6H50CH3' -OCH3 does not sufficiently increase electron density on the ring. Recall that C6"50H undergoes coupling in weakly
alkaline medium which converts C6HsOH to the more reactive CtJJ.sO-. In options (b) and (c), presence of electron-withdrawing-N02
groups increases electrophilic character to such an extent that these diazonium cations can couple even with the compounds having
weak electron-releasing groups. Option ( d) undergoes coupling reaction easily because -NMe2 is sufficiently electron-releasing.
Recall that the -Cl group present in the o- and p- positions to the electron-withdrawing group is activated toward nucleophilic
substitution, hence only -Cl present on the o- and/ or p-position to the N02 group will be replaced.
In Libermann's nitroso test 2° amine and phenol are used as reagents.
In p-aminophenol all the four positions (2, 3, 5 and 6) can be coupled (positions 2 and 6 in presence of tt+ and positions 3 and 5 in
presence of OH-).
-
6.
0 groups.
CH3
�
The above reaction is an example of Hofmann Martius rearrangement.
Theoretical question.
>lCORRECT
1�
=
�.HO
NHCHJIO
20.2
2.
3.
II
1° Amine, e pair
delocalized
m
2° Amine, e pair
delocalized
H
II has maximum electrons density on N and it is localized; in N electrons are delocalized due to two -C
N(CH3}zHO
17.
�
I
1° Amine, e pair
N
2° Amine, e
6
>
>
�N
-0- a
p-Chloroaniline
(-0 is non-reactive)
It reacts with AgN03. The
(-0 is present
as
reactive)
three other options (a, b and d) are not proper reagents to distinguish.
1.
(a)
(e)
2.
�
(c)
r
!WJ
I�
!WJ
(a) All the three produce soluble am monium salts.
C6"sNH2.HQ, C6"sN+"2CH3a-, and C�H(cH3>:zO.
(b) All the three produce same quaternary ammonium salt, C6"5N+Me31-, of
(c) C6HsN"2
c:=
course
by taking different amount of CH3I.
0
C�
� - �--C6H5 Neutral( duetodelocalizationofeonN to--C=Og roup),
insoluble in dil acidor base
Neutral, insoluble in acidor base
CH3
I
�- CH3 ----+ No reaction.
C6H5
Basic,
soluble in acid, insoluble in base
Addie (Soluble in base)
Neutral
Soluble
.
(Insoluble in acid or base)
CH3
I
C6"5 N--CH3
------+
No reaction ; Basic (insoluble in bas e, soluble in acid)
(e) All the three undergo ver y fast ring bromination to.give corresponding tribromop roducts.
3.
•
�OH
(<)©
4.
(a) C6"sNHCOC6"s + NaOH ·� C6"sN"2 + C6"scoo-Na+
(b)C6H5NH2
+
0
C6H5NH-C-H +CH30H
�O)CH,
lSJ.(NOz + H,O
0
II
H3CO-C-H
ll
�
Methyl formate
Aniline
+
(d)
3HBr
(-N� is more activating than-cH3)
(<)
I:,
AoNI-Ii
+
2Br2(aq.)
3
�
(f)
3
(Serond-NOz is not introdured
(-NHCOCH3 is more
electron-releasing than -CH3)
because the ring is deactivated
due to -N02 group already present)
,
Deactivated
Activated
Br2/Fe
-0-- H �-0
. Br� N-C �
+
o-Isomer
Benzanilide
(Ring having -NH-part is
activated while other having
--CO- is deactivated
(b)
COCl2 + C6HsNH2
[E] and [F]
(d
)
�"'
?
[I]
(<)
~+
[K]
'""'°"""
[L]
1111
A nitrile has the formula R-C = N: (or R-CN). The carbon and the nitrogen of a nitrile are sp hybridised. In
IUPAC systematic nomenclature, acyclic nitriles are named by adding the suffix nitrile to the name of the corresponding
hydrocarbon, the carbon atom of the -C
2
1
=
N group is assigned number 1.
2
3
CH3-C::N:
5
1
CH2 =CH-C::N:
Ethanenitrile
Propenenitrile
(acetonitrile)
(acrylonitrile)
4
3
2
1
ffi2 = Oiffi2a-I2-C = N:
4-Pentenenitrile
C yclic nitriles are named by adding the suffix carbonitrile to the name of the ring system to which the -CN group is
attached.
Benzenecarbonitrile
Cyclohexanecarbonitrile
(benzonitrile)
Common name of nitriles (given in parentheses) are derived by replacing the -ic acid or -oic acid ending of the
corresponding carboxylic acid by -onitrile. Alternatively, nitriles are sometimes given radkofunctional names as alkyl
cyanides.
Isonitriles are functional isomers of nitriles; the latter being more stable than the former.
EB
9
R-N=C:
Isonitriles
heat
Nitriles
These are commonly called by their common names, in which the prefix iso is added before the name of the
isomeric nitrile. In IUPAC nomenclature, these are named as alkylcarbylamines.
EB
ED
0
H
Propionoisonitrile
Acetoisonitrile
1.
Give IUPAC names for the following:
(c)
1.
9
CH3CH2-N = C
Ethylcarbylamine
Ethyl isocyanide
Ethyl isonitrile
C 3-N a C
Methylcarbylamine
Methyl isocyanide
Methyl isonitrile
T820Ia
0-0ICN
From alkyl halides. Aliphatic nitriles are prepared by treating alkyl halides with sodium cyanide in a solvent
that dissolves both reactants ; in dimethyl sulphoxide (DMSO), reaction occurs rapidly and exothermically at
room temperature.
R-X
1° or 2°
+
CN-�R-CN
+
x-
alkyl halide
Remember that this reaction is of the SN2 type and is limited to 1° and 2° alkyl halides ; 3° alkyl halides undergo
exclusively elimination reaction because CN- is a strong base (recall that HCN, conjugate acid of CN- is a very
weak acid).
NaCN
o-�2Cl
Q--cH2CN
DMSO
Benzyl chloride
.0-a
I
Benzyl cyanide
NaCN)
DMSO
N
0-c
Cyclopentyl chloride
Cyclopenyl cyanide
In case silver cyanide is used in place of potassium cyanide, isonitrile is the main product, while nitrile is formed
· .
only in small amount.
CH3I +AgCN �
·
�NC +Agl
Methylcarbylamine
Remember that aryl and vinyl halides are unreactive toward nucleophilic substitution reaction, hence aryl
nitriles and vinyl nitriles can't be prepared by this method.
C6H5Cl
2.
From arenediazonium salts. Aryl
or
CH2
•
CHO
�
DMSO
No reaction
nitriles are prepared from diazonium salts by
involves the reaction of aryl diazonium salt witli cuprous cyanide.
Ar-N2+ x�
Aryl diazonium halide
Ar--CN+N2
Sandmeyer reaction
which
By dehydration of amides. Amides react wi th P4010 (a compound that is often called phosphorus pentoxide
and written P205) or with boiling acetic anhydride to form nitriles.
P4010 or (CH3C0)20
-------+
heat,
(R
=
(-H20)
RCN + H3P04
or
CH3COOH
alkyl or aryl group)
This method is useful for preparing nitriles which can't be prepared by nucleophilic substitution reaction
between alkyl halides and cyanide ions.
4.
By the dehydration of aldoximes with acetic anhydride or P4010
(CH3C0)20
RCH=NOH
RC=N+IfiO
Nitrile
Aldoxime
(R may be alkyl or aryl)
5.
From Grignard reagents.
RMgX
----+ RCN+Mg(O)X
QCN
+
Cyanogen chloride
6.
Carbylamine reaction
(only for isonitriles).
C2H5NH2 + CHQ3 + 3KOH ---+ C2H5NC + 3KO + 3H20
1.
1.
2.
(i) CH3CN, and (ii) CH3CH2CN
(a)
Try to convert ethanol to
(b)
Give steps involved in the synthesis of Me3CCN from Me3CCOOH.
Nitriles are more polar than isonitriles, hence they have high dipole moment, high b.p. and are more soluble in
water. Recall that carbylamines (isonitriles) are extremely unpleasant, while nitriles are pleasant smelling.
Hydrolysis. Since nitriles are converted to carboxylic acids on hydrolysis,
derivatives.
these are classified as carboxylic acid
Like amides, nitriles when heated in aqueous acid or base for several hours give carboxylic acids. Further
like amide hydrolysis, nitrile hydrolysis is irreversible.
R-C
=
R-C
=
N + H20
N
H20
+
H3o+
+
+
OH-
RCOOH + NH4+
RCOO-
---+
+ NH3
Mechanism for acidic hydrolysis of nitriles
H
R-C:N: + H30
('\+
R-C=NH
+
Protonation of the nitrile
:o-H
I
..
R-C=NH
Iminoacid
(Amid e tautomer)
+
R-C=NH
••
+---+
Protonated nitrile
(b:_H
�
R
-l
=NH
(+ve change on N makes C more electron-deficient)
.
�
HO+
.
[
:Q-H
I
•
R-C
•
+
+·· ]
0-H
=NI\
+---+
fl
••
R-C-NI\
Protonated amide
0
II
.
+
R-C-NH2 + H30
Amide
•
In presence of concentrated H2SO4' the reaction stops at the protonated amide which constitutes a useful way for
preparing amides from nitriles. In presence of dilute -acids, amides undergo further hydrolysis to form carboxylic
acids as the final product.
+··
,.oH
Q:
II ..
R-C-NH2
'-II
..
I
+
I
..
I
R-C-OH
I ..
.•
R--C-OH2
R-C-�
!':lli2
QjH3
Ammonium ion
Oxoniumion
(a proton is lost from
0 and gained at N)
Protonated amide
Amide
�OH
:OH
+··
OH
II
.. H
R-C-O
Protonated carboxylic acid
Mechanism for basic hydrolysis of nitriles
0
OH
II
I ..
R-C=NH
R
Hydroxy imine
/
+
c
'o_
Here also, amides can be isolated under appropriate (milder) conditions.
3.
carbon-nitrogen triple bond of nitriles is much less reactive toward
nucleophilic addition than is the carbon-oxygen double bond of aldehydes and ketones, strongly basic
nucleophiles such as Grignard reagents and organolithium compounds easily add on nitriles to form ketones
Addition of Grignard reagents. Although the
R'
R-C = N + R'MgX
(i)diethylether
(ii)H20
R-CI = NH �
Ho+
heat
R'
=
I
R-C=O
R'
I
I
R-C
R'
R-C=O +NH/+ u+
N + R'Li --+ R-C = N-Li +
Since nitriles are more reactive than ketones, the next molecule of R'MgX
will add to fresh molecule of RC = N to
form ketone rather than with ketone to form a 3° alcohol. Although a nitrile has a triple bond, addition of the
Grignard or lithium reagent takes place only once because if the addition takes place twice, nitrogen will get a
double negative charge which is not possible.
R'
R'Li
I
R-C = N ---+ R-C = N-Li +
R'Li
I
R'
�
R-C-N2- 2u+
I
R'
(The dianion does not form)
1.
Compare the reaction of Grignard reagent with an ester and a nitrile, explain the difference between the intermediates formed in two
cases when one molecule of the reagent adds on the compound (ester or nitrile).
Reduction of nitriles. Nitriles can be reduced to 1° amines
R-C
=
N
!>Y catalytic hydrogenation or LiAlH4.
2H2, Raney Ni, 140°C
;;..._---'---- -t
--
R--CH2NH2
Nitriles can also be reduced to aldehydes by means disobutylaluminium hydride (DIBAL-H). lhis reaction is
also known as Stephen reduction.
R-C
5
(iso-Bu)2 Alli
N
� CEN
s.
�o
(i) (iso - Bu).iAlH
(ii)�O
H
Condensation with aldehydes. The a-hydrogens of nitriles are appreciably acidic, but less than those of aldehydes
5
and ketones. The acidity constant for acetonitrile (CH3CN) is about 10-2 (pK11 = 25). Other nitriles with
a-hydrogens show comparable acidities, and consequently these nitriles undergo aldol type of condensations.
0
II
c6H5 CH + C6H5CH2CN
(a)
C2Hso-
-----+
C2H50H
C�5CH = cCN
l
C6 Hs
The note-worthy properties of alkyl isonitriles are hydrolysis, reduction and addition reactions.
�504 or HO) to
Alkyl isonitriles are hydrolysed only by dilute mineral acid
form 1° amines
-�
(b)
lsonitriles do not undergo basic hydrolysis because of their inability to undergo the attack by OH- ions.
Catalytic hydrogenation or reduction with lithiumaluminium hydride gives 2° amines.
R-N-+C
(c)
LiAlH
---4
RNHCH3
lsonitriles undergo certain addition reactions in which addition takes place only on carbon.
RN•C=O
RN�C
1/8Ss
Alkyl isothiocyanate
Alkyl isonitrile
Alkyl isocyanate•
RN .. C=S
�
These are functional isomers, e.g.
o:
#
:?"
N
R-
Alkyl nitrite
"'-··­
g:
Nitroalkane
Nitromethane, used as a high-energy fuel for race cars, is a liquid with a b.p. of 101°C, while methyl nitrite is a
gas (b.p. -12°C) which causes dilation of blood vessels, on inhalation, hence it is used for the treatment of angina pectoris
(a heart disease).
HQwever, remember that alkyl nitrites are esters of nitrous acid (HONO) while nitro compounds are not esters
because these can neither be prepared by the interaction of alcohol with an acid (HONO) nor they undergo hydrolysis to
form alcohol and acid (nitrous acid). Actually, nitro compounds are regarded as nitro derivatives of the hydrocarbons.
..
Recall that methyl isocyanatf>.
MlC was
responsit:>le for Bhopal tragedy in Dec.
1984 .
1.
By
the action of nitrous acid on alcohol.
NaN�,W
C:zHsONO +Hp
(+HONO)
C2H5OH
.
2.
By
the action of nitrogen trioxide
(N203)
I
on.alcohols.
2C2fisOH + N203----+ 2C:ifisONO +H20
Nitrogen trioxide is generated in situ i.e. it is prepared by the action of cone.
HN03 on a mixture of cone. H2S04
and alcohol and then adding copper turnings.
3.
By
the action of alkyl iodide and pbtassium nitrite.
C2fisI + I<N02 ----+ CiHsONO +KI
Properties. Its two worth-mentioning properties are hydrolysis and reduction.
C:iHs--0-N
•
W, OH-, or, neutral medium
0 --------+ C2fis--OH +HONO
Nitro group is found to be a resonance hybrid of the following two structures as evidenced by its equivalent
bond length of the two N---0 bonds.
,,
5:
+/··
+.J'J-
-N,
�
0
-N
�g:
Resonance hybrid of the-N� group
Resonating structures of the- N� group
Preparation.
1.
By heating an alkyl halide with of solution of silver
nitrite-;
C2H5Br + AgN02 ----+ CiHsN02 +AgBr
Some alkyl nitrite is also formed as a side-product ; however, the two compounds can be separated easily by
fractional distillation.
2.
By
the vapour phase nitration of alkanes.
CH3CH3 +
HN03 (fuming)
400°C
--+
CH3CHiN02
+ �N02
The mixture of nitralkanes can be separated by fractional distillation.
\
3. 3° Nitroalkanes can be prepared by the oxidation of the corresponding 3°
� with I<Mn04.
KMn04
--+
Properties. Presence of positive charge on N makes
-N02
a powerful electron-withdrawing group which
makes a-hydrogen atom acidic, much more than those of aldehydes and ketones. Further, the carbanion formed by the
removal of a-H is stabilized due to resonance
(see
compared to 3° nitroalkanes having no a-hydrogen.
/
acidic character). Hence 1° and 2° nitroalkanes are very reactive as
Acidic nature. Primary and secondary nitroalkanes dissolve in aq. NaOH to form salts.
aq.NaOH
(-�O)
(b)
Action of alkaline halogen (X2 + NaOH)
C2, NaOH
H3CNo2
(Otloropicrin),
Nitrochloroform
a lachrymatory substance
Nitromethane
(c)
Action of nitrous acid.
NO-Na+
NOH
(i)
RCH2N02
A 1°
II.
HONO
RCN02
------+
R CN02
------+
Sodium nitrolate (red)
Nitrolic acid
nitroalkane
II
NaOH
N=O
(ii)
�CHN02
A 2°
(iii)
HONO
------+
NaOH
No reaction
------+
Pseudonitrol (blue)
nitroalkane
�CN02
I
�CN02
NaOH
------+
No reaction due to absence of a-H
A 3° nitroalkane
Recall that this difference of nitroalkanes toward nitrous acid forms the basis of the Victor Meyer test for
distinction of 1°, 2° and 3° alcohols.
(d)
Condensation with aldehydes and ketones.
aldehydes and ketones.
Nitroalkanes having a-H undergo base catalyzed
1.
condensation with
.
�·
l!fTJ
the
.
"·_'
.
,.
Give the starting aldehyde I ketone, and nitroalkane used for synthesizing each of the following compounds.
CH3
(a)
(e)
C6H5CH
I
=
CN02
Mannich reaciion. This is the condensation between formaldehyde and compound having
or a.1° or a 2° amine to form Mannich base.
ammonia
·
N02
HQ
I
active hydrogen and
R'
I
�CHN02 + HCHO + R'NH2 ------+ �CCH2 NH + H20
2.
Reduction
(a)
Reduction in acidic medium, like Sn + HO, Fe + CH3COOH ; LiAIH4; or
H2-Ni, gives amines.
Sn+HC
(b)
Reduction in neutral medium, like Zn dust
+
NH40 gives hydroxylamines.
Hydrolysis. Primary nitroalkanes are hydrolysed to hydroxylamine. and carboxylic acid, while secondary
nitroalkanes are hydrolysed to ketones.
HO
2 (CH3)iCHN02 --+ 2(CH3hCO + N20 + H20
Tertiary amines do not undergo hydrolysis.
1.
Direct nitration. This is the most important method for preparing aromatic nitro compounds. Nitration can be
done by variety of reagents, like a mixture of cone. HN03 and cone. �SO41 acetyl nitrate (N205 in acetic anhydride),
fuming HN03 etc. Since -N02 group is deactivating, it is very difficult to introduce the second and third -N02
group, however this can be done by carrying out the reaction, at high temperature and with strong nitrating
agent like fuming HN03. For example,
(i)
&
NCJi
©
Conc.HN�
Cone. H25()4' 100°c
N�
(ii)
02N
Cone.
H;r5CJ4' 60"C
Nitrobenzene
Benzene
m -Dinitrobemene
A
Cone. 825()4'Conc. HNO:i
fumingHN03
N02
6
Conc.HN03
lOO"C
IOO°C
1, 3, 5-Trinitrobenzene
Benzene
&
m
NO2
-Dinitrobenzene
Recall that nitration is an electrophilic substitution hence presence of electron-releasing group like
NH21 --CH31 --OR etc., in the nucleus facilitates nitration. Thus aromatic compounds bearing these grou
toluene,
aniline,
phenol etc.) can be nitrated easily as compared to benzene.
.
.
2.
(i)
Q
Q
+
NaN02
N02
p-Nitroaniline
·
-
NaN02
HO
CH3
(ii)
�a
From diazonium salts (indirect method). Since diazonium ion can be easily replaced by -N02 g oup,
arenediazonium salts are used for preparing the aromatic nitro compounds which cannot be prepared by
ct
method.
Cu heat
c�
+
NaN02
HBF4
�
-
��.
NOi
9
9
NaN02
Cu heat
N 02
p-Dinitrobenzene
+
HCl +
N2
IE.
1.
.
.
.
Give steps involved in the conversion of aniline to o-dinitrobenzene.
Nitro compounds are highly coloured (yellow, orange, red etc.), insoluble in water. Its three worth-mentioning
properties are reduction, electrophilic substitution and nucleophilic substitution.
1.
Reduction. Aromatic nitro compounds can be reduced to a variety of products through the following sequence.
Nitrobenzene
Aniline
Phenylhydroxylamine
Nitrosobenzene
y
11
�HsN
-----+
Azoxybenzene
�lfsNH
�lfsN
C6H5N-+0
II
�lfsN
-----+
Azobenzene
I
�lfsNH
Hydrazobenzene
Each of the above product can be isolated depending upon the nature of the reducing agent. Following table lists
the important products along with the corresponding
l'llducing
agent for that very particular compound..
Reduction of C6H5N02 under different conditions
Rcugc11t
1.
Product
Reduction in addic medium
Sn+ HO, Zn+ HO, 5n02+ HO, Fe+ H20 + HO
2.
Reduction in neutral medium
Zn dust + NH4Cl, Al-Hg couple + H20
3.
(b) Zn+ NaOH, CH30H
(c) Zn + NaOH; CiH50H
s.
Phenylhydroxylamine
Reduction in alkaline medium
(a) Na3As03 + NaOH
4.
Aniline
Azoxybenzene
Azobenzene
Hydrazobenzene
Reduction by metal hydrides and cata -�
H2+ Pt or Raney Ni ; or LiAIH4
Aniline
Electrolytic reduction
p-Aminophenol
in presence of dil.�4
In case the compound contains two -N02 groups, selective reduction (reduction of only one -N02 group) can
be achieV'ed by using (NH4)i5 or NaiS.
m
&
N�
-Dinitrobenzene
o/�
lmolSn�
NOi
2.
�
HO
c6l
(NH4h5
orNa:zS
o/
m
N�
N02
o/
-Nitroaniline
polysulphide
warm
N�
NH2
N02
Substitution is benzene nucleus of nitrobenzene. Resonance in nitrobenzene imparts following characteristic
features.
3+
Resonance hybrid
of nitrobenzene
Resonating structures of nitrobenzene
(i)
It imparts partial double bond character to carbon-nitrogen bond causing replacement of -N02 a difficult task.
However, if second -N02 group in present in the ortho- and para positions, one of them can be replaced by a
nucleophile.
Nu
¢
N�
. (Where, .Nu -OH,� or oc;Hs)
(ii)
It imparts partial positive (especially o- and p-positions) charge on the ortho. and para positions deactivating
benzene nucleus for electrophiles and making the ortho- and para- positions reactive for micleophiles. Thus
nitrobenzene undergoes electrophilic substitution with difficulty and at m-position, however it undergoes
nucleophilic substitution at ortho- and para-positions easily.
(a)
As mentioned above, nitrobenzene undergoes electrophilic substitution only
and at meta position. However, nitrobenzene having activating group like -R, OR, -NH2, etc. undergoes electrophilic substitutions relatively easily and at the position governed by the
activating group rather than the nitro group.
Electrophilic substitution.
under dra5tic conditions,
�
�
�
N�
N�
2, 4, 6-Trinitrotoluene
(TNT)
-'
Nucleophilic substitution. Recall that benzene is inert to nucleophiles ; however nitrobenzene undergoes
&
nucleophilic substitution in o- and and p-positions, when fused with KOH.
©
KOH
----+
heat
Nitrobenzene
-
�
OH
+
OH
o-Nitrophenol
p-Nitrophenol
When nitrobenzene having electron-withdrawing group in o- or p-position is heated with ethanolic
KCN-the -N02 group is eliminated and a -COOH group is introduced in the ortho position to the
leaving-N02 group (Von Richter reaction).
�
&fu
KCN,<;HsOH
Br
¥
H
Br
KCN,<;fisOH
HOOC
"©(r
0-13
(a)
I
C H5N-OH
�
C6 H5NHOH
C6H5NO
[A] formed by
[BJ formed
[C] Remember that N
neutral reduction
by oxi. of hydroxylamine
behaves like C
-
=
C6HsN
=
NC6Hs
[D] Reduction of
0 group
nitrosobenzene
0 group
(b)
[F], Formed by Kolbe's reaction
[E], Formed by electrolytic reduction -
0
II
(c)
OC'cooH
KOH
�
�
C2H50H
NHi
2
(d)
�o
H
Formed by intramolecular redox reaction
CH3
�
�
�vc::��
N�
N02
�02
Three-N02 groups in o-andp-positions
make -CH3 group reactive
(acidic)
Br2, Fe
(e)
&�
-CH3 is activating group
[HJ and [I]
cis- and trans-isomers
��
Br
[j]
KCN,
HOOC��
C�1PH
Br
[kl Formed by
Von Richter reaction
(f)
�
N 02
-N02 group makes
�H3 reactive
c,H,ONo
(a)
N02
Although m-nitrobenzoic acid can be prepared easily by oxidising m-nitrotoluene, but it is very difficult to
convert nitrobenzene to m-nitrotoluene because nitrobenzene does not undergo Friedel-Craft reaction. Hence
indirect route is adopted, it is advisable to proceed backward for getting different steps.
&�
COOH
x
�
CN
(i)HONO
(ii)CuCN, HCN
.
&
Fe
Br
(b)
-N02 group in the m-position to -CH3 i.e. in the o-position to the existing -N02 group can be introduced
easily by converting existing -N02 group (m-directing) to the -NH2 group (o, p-directing).
9
Ntti
¢� �c:
�
�
N
N
�c: � Qf y
Sn/HCI
CH3
¢
(CH3CO)p
CH3
HN03
CH3
C1i.3
Now the -NHCOCH3 group can be easily removed via diazotisation.
+
H
�
Cli.3
N�
HONO
0°C
CH3
�fisOH
orH3ro2
C1i.3
C1i.3
___, Br
(c)
V
Br
Br
Remove -N02 group already present in the molecule (as in the above example) through diazotisation but by
first introducing three Br.
©
Br2
�/Pt
Br
Water
*
Br
Br
(i)HONO
(ii)H3P02
v
Br
Br
Br
Solution:
Let us summarise the given reactions.
A
reduction
B
�
D
c
(haloform reaction)
NaOH
E
Formation of D from E indicates that E (CH3N02) is a nitromethane and hence D is trichloronitromethane,
nitrochloroform or chloropicrin. Hence compound C should be chlorofo�m, CHC13 consequently compound B
should be CH3CH20H and A as ethyl nitrite.
CzH50N
=
0
reduction
---+
C2H50H
(A)
a2,mr
(B)
21.1 (MCQ
1.
2.
-
(C)
Cyclohexanecarbonitrile
(c)
Cyclohexanonitrile
(b)
(d)
(D)
3.
(a)
(c)
R-C-NH2
Both
(b)
(d)
Methyl nitrite shows tau tomerism
0-CN
Cyclohexanenitrile
I
None of the three.
Predict their relative acidity.
Protonation of acid amides (RCONH2) is likely to give
+oH
II
(E)
ONE option correct)
Which of the following is not the correct name for
(a)
CH03
0
II
(a)
(b)
(c)
(d)
+
R-C-NH3
None.
4.
Both are equal in acidic character
I is more acidic than II
II is more acidic than I
None of then is acidic.
The o r de r of inc reasin g boiling points amo ng the th ree
compow1ds is
CzH50NO(I),
C2H5NOz(II)
CzH6(III)
(b)
(c)
s.
m
<
I
<
n
(d)
Whichof the following gives a primary alcohol on reduction ?
(a) Alkyl carbylamine
(b) Alkyl nitrite
(c) Nitroalkane
(d)
None ofthe three.
When a nitrogenous substance X is treated with nitrous acid, a
·
6.
(a )
blue colour is obtained. The compound X can be
(a)
(c)
CH3CH2N02
CH3�0NO
(d)
(CH3hCHN02
(CH3CH)20NO.
The compounds X and Yin the following reaction respectively
7.
·
(b)
are
(a )
(b)
(c)
g-NHa+0o-�
0--NH
-00--NH-HN-O
-o--o0--NH-HN-O
ando-�-�-0
(d)
LNOi
u
Conc.HN03
Conc.HiSQ4
.
(i)N32Cr20,/H
and
OH and
and
8.
(b)
H0
�N
+
(ii) Soda lime
�
(c)
Both are equally good
(a)
©
(b)
&N
(d)
None is suitable.
(d)
A neutral benzenoid nitrogenous organic compound has zero
dipole moment, its probable structure is
9.
�
Which of the following method gives good yield for
1, 3, 5-
(c)
trinitrobenzene, commonly known as TNB?
02
©'
QNOi
21.2 (MCQ 1 or >1 option correct, Passage based, Matching, AIR)
2.
DIRECTIONS for Q.1toQ.9: Multiple choice questions with one
or more
1.
than One curred option(&).
(a)
6
QrN�
NOi
(b)
©
a
(c}
Which of the following shows tautomerism ?
(a)
C�50H
(b)
CH3N02
(c}
Which of the following reaction can forma nitrile when treated
.with NaCN in DMSO?
a
·
(d)
�
A
(d)
. (CH�)JCN02
Which of the following reacts with nitrous acid ?
(a) Acetamide
(b) 2-Nitrobutane
(c)
2-Methyl-2-nitropropane
4.
Which of the followingcompounds canredu<E Tollen's reagent?
(a)
CH3CHO
(b) C6HsN02
.
(d)
CH20HCOCH3
(c) <;HsNffOH
s.
p-Nitroaniline can be obtained by
(d)
N�
<;HsCH�02
3.
(a }
Diethylamine.
(i)HN03
(ii)�S04
6
(c)
c?
(d)
(i) (CH3CO)i0
(ii) HN03/ffiS041
(iii)H3d
10.
6.
7.
8.
9.
(A)
(B)
(C)
HN03
(D)
o,
+
o-
,o
' +/
N
(a)
¢
°'
14.
15.
0
Q
�+;
N
(c)
13.
0
0
0
o-
0
o-
N
¢
o-
(b)
(c)
(d)
(a)
(b)
(c)
(d)
1°amine
2°amine
Hydrolysis
Reductive amination
Column-II
Yellow precipitate
Acetoacetic ester
1-Butyne
Lemon yellow oily liquid
alkylation.
+;
.
(d)
_
Column-I
Bromine water
Nitrobenzene
CID3
Acidic hydrogens
(a)
Statement-1: Alkyl isocyanides with acidified water gives alkyl
formamide.
Statement-2 : In isocyanides, carbon first acts as a nucleophile
and then as an electrophile.
Statement-1: Methyl isocyanide reacts with ozone to form methyl
isocyanate.
Statement-2 : Methyl isocyanate was responsible for Bhopal
tragedy.
Statement-1 : Alkyl cyanide can be prepared by carbylamine
reaction
Statement-2 : Ethyl amine when heated with chloroform in
presence of alcoholic KOH, cyanide is formed.
Statement-1 : Nitrobenzene does not undergo Friedel Craft
12.
N
(b)
Nitriles
lsonitriles
Ketones
Phthalimide
Instructions for Q. 12 to Q. 16 : Following questions are Assertion
and Reasoning Type Questions :
Note : Each question contains STATEMENT-1 (Assertion) and
STATEMENT-2 (Reason). Each question has 5 choices (a), (b), (c),
(d) and (e) out of which ONLY ONE is correct.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct
explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a
correct explanation for Statement-1.
(c) Statement -1 is True, Statement-2 is False.
(d) Statement -1 is False, Statement-2 is True.
(e) Statement -1 is False, Statement-2 is False.
The products of reaction of alcoholic silver nitrite with ethyl
bromide are
(b) ethene
(a) ethane
(d) ethylnitrite
(c) nitroethane
Which gases are poisonous?
(b) Mustard
(a) Lewisite
(d) MIC
(c)
Phosgene
Butanonitrile may be prepared by heating
(a)Propyl alcohol with KCN
(b) Propyl magnesium chloride with Cyanogen chloride
(c)Butyl chloride with KCN
(d)Propyl chloride with KCN
The
representations
of
resonance
structures
of
p-nitrophenoxide ion are
-
(D)
11.
S03H
Column-II
Column-I
(A)
(B)
(C)
Statement-2 : Nitrobenzene is used as solvent in laboratory and
industry.
Statement-1 : Alkyl isocyanides in acidified water give alkyl
formamides.
Statement-2 : In isocyanides, carbon first acts as a nucleophile
and then as an electrophile.
16.
0
LUTI
I
-��-
Jfi/ "'
� ,,.., •11.
·
jpgJ,j1,R
··�
1.
1.
(a) 2-Methylpropanenitrile
(a)
(b)
(b) Ethylcarbylamine.
(i)SOCl2
Na2Ci:z07, H20
(i)
CH3CH20H
(ii)
CH3CH20H
Me3CCOOH
H2S04,heat
PBl'J orHBr
SOCl2
CH3COOH
CH3CH2Br
(ii)NH3
NaCN
CH3CH2CN
(l)NH3
Me3CCOCI
n
u P.4010• heat
CH3CONH2
Me3CQ.{.
(c) 2-Phenylbutanenitrile
P4010
CH3CN
.
,
.
�Wt·�,·
liir!!�:�� -,
.
,
,
.,,__
1.
.
-""'
Esters react with Grignard reagent to form first-ketones and then 3° alcohols.
0
0
II
CH3-C-OC2H5
CH3Mgl
--�
lf�
I
f
CH3- -OC2H5
�
I
-Mgl(OC2Hs)
-----+
CH3- CH3
CH3MgI
lf�
I
----t
f
OH
CH3-C-CH3
�
CH3
CH3
I
H+
CH3- -CH3
I
CH3
(More reactive than ester)
However, alkyl nitriles react with Grignard reagents to form ketones as the final product even in presence of excess of Grignard reagent.
N-(Mgl)+
R-C=N
�
-Mgl(CH3)
R-C-CH3
R-C=N
(Regenerated)
I.
l
H+
0
II
R-C-CH3
Unlike, the ester intermediate I, where loss of negative charge by elimination of OC2H5 leads to ketone loss of negative charge by
eliminating CH3 from the nitrile intermediate Ia merely reverses the reaction to form nitrile. Hence the intermediate la undergoes
hydrolysis to form ketone which, being less reactive than nitrile, does not react with the next molecule of Grignard reagent.
1.
1.
6
Aniline
CH3COCI
protection of
-NH2group
6='
&=� +¢='
Conc.HN03
r&
&�
ortho
Acetanilide
0
H30+
�
o-Nitrobllnzene
&
+
N�
NaN02
HBF4
Separation
by fractional
crystallization
N02
para-
-
N�
NaN0 2
Cu, heat
&
NO,
o-Dinitrobenzene
1
6
2
7
3
8
4
9
s
1.
Cyclohexanenitrile is the wrong name.
2.
The cation produced by the protonation of carbonyl group of amides is stabilized by resonance.
(OH
..+
:o
II
R-C-NH3
II�.
H+
+
R-C-NH2
+-----
An acylammonium ion
Acid amide
(the positive charge is
localized on N)
:oH
+------+
I
+
R-C•NH2
o-protoriated amide
(the positive charge is delocalized)
3.
The conjugate base of the aci-form
4.
Nitro compounds, being relatively more polar than nitrites, have higher boiling points than the corresponding nitrites.
s.
R---0-N
7.
Reduction of nitrobenzene in alkaline medium gives hydrazobenzene which rearranges in presence of HO to form 4, 4'-diaminodiphenyl,
8.
Introduction of one-N02 group in benzene nucleus deactivates for further nitration. However, in option (b), nitration of o-nitrotoluene
is not difficult because of the presence of activating-CH3 group. Hence 2, 4, 6-trinitrotoluene is easily prepared from o-nitrotoluene.
Further, -CH3 group is easily oxidizable when-N02 groups are present in benzene nucleus as compared to oxidation of C6H5CH3.
9.
Neutral nature of the compound indicates that it should be a nitro compound. Further zero dipole moment indicates its symmetrical
structure, i.e. it should be a benzene derivative having same group (-N02) in positions p- to each other.
=
0
reduction
--4
R---OH
+
(II) is more stable.
H2NOH
commonly known as benzidine (benzidine rearrangement).
21.2
>1 CORRECT
OPTION
MATCH THE
FOLLOWING
AIR
1.
C6H5CH20
1
b,c
2
6
c, d
7
a
13
s
c,d
16
a
10
11
12
(b) is arylsubstituted alkyl halide, hence it undergoes nucleophilic substitution; while in (c)
Cl is activated by the
presence of electron-withdrawing -N02 groups in ortho- and para-positions.
2.
tert-Nitroalkanes do not have labile H (H in a-position to the -N02 group) and hence can't exhibit tautomerism.
3.
tert-Nitro compounds (Me3CN02) do not react with HONO because they do not have any a-H. The three others reach with HONO as
usual.
4.
Aldehydes, a-hydroxyketones and hydroxylamines reduce Tollens' reagent.
5.
-S03H group present in o- and p-positions are easily replaced.
·
2° Nitroalkane
Pseudonitrol (blue).
1111
·'